I did pi*Radius^2 x 2 for the circles, then pi*D to get the circumference then multiplied that to get the area of the rectangle on the cylinder and added the 2 values together to get the whole area. Area was around 22+? Then i did 5x7=35 for how much paint he had. Then i said he had enough paint.

Same! But apparently it's wrong

Reply 248

GuruOfMaths

11

wait what was the quadratic formula question like

Original post by nxke.lucas

2n^2-3

Reply 249

lucayah

7

Original post by Olliec01

U said u got X< 2.5 and X< 1.6

i mean x<0.4 and x<1.6 i couldnt exactly remember my answer but i recalculated it doing what i did in the exam thats what i got

Reply 250

Eli0038

6

For the standard form question would 1.3 x 10^2 be correct or is it only 130 seconds

Reply 251

Truzerx

8

Original post by lucayah

i mean x<0.4 and x<1.6 i couldnt exactly remember my answer but i recalculated it doing what i did in the exam thats what i got

what question was this I cant quite remember

Reply 252

zimple

7

for 1st part angle at A inside right angled triangle is 90-37 = 53 so other angle in same triangle at B is 180 -90 -53 = 37
frighta ngled triangle with C gives angle at B as 30 degrees
so angle aABC is 37 + 30 = 67

for second part use cosine rule
a = 8 km b = 9km angle is 67 degrees
gives c = root 88.73 = 9.4 km

for third part find two North south sides of the same two triangles in part 1
Both use sin = opp/hyp
1st gives sin 53 = opp /8 so opp = 6.389
2nd gives sin 60 = opp / 9 so opp = 7.794
difference between two = 1.405 is Northern dispalcemt of A from C
so use sin again to find angle opp = 1.405 and hyp = 9.4
gives sin = .01495 so angle is 8.6 degrees
add on 270 for true bearing = 278.6

Original post by jacobsuresh

can some write their working out for the bearing question i dont remember my answer but i remember doing cosine rule and 0.5absinC

Reply 253

Eeieoeoeieii

4

Tf did you get sohcahtoa from lmfao these aren't right angled triangles

Original post by zimple

for 1st part angle at A inside right angled triangle is 90-37 = 53 so other angle in same triangle at B is 180 -90 -53 = 37
frighta ngled triangle with C gives angle at B as 30 degrees
so angle aABC is 37 + 30 = 67

for second part use cosine rule
a = 8 km b = 9km angle is 67 degrees
gives c = root 88.73 = 9.4 km

for third part find two North south sides of the same two triangles in part 1
Both use sin = opp/hyp
1st gives sin 53 = opp /8 so opp = 6.389
2nd gives sin 60 = opp / 9 so opp = 7.794
difference between two = 1.405 is Northern dispalcemt of A from C
so use sin again to find angle opp = 1.405 and hyp = 9.4
gives sin = .01495 so angle is 8.6 degrees
add on 270 for true bearing = 278.6

Reply 254

RebelMaster7

1

Original post by zimple

for 1st part angle at A inside right angled triangle is 90-37 = 53 so other angle in same triangle at B is 180 -90 -53 = 37
frighta ngled triangle with C gives angle at B as 30 degrees
so angle aABC is 37 + 30 = 67

for second part use cosine rule
a = 8 km b = 9km angle is 67 degrees
gives c = root 88.73 = 9.4 km

for third part find two North south sides of the same two triangles in part 1
Both use sin = opp/hyp
1st gives sin 53 = opp /8 so opp = 6.389
2nd gives sin 60 = opp / 9 so opp = 7.794
difference between two = 1.405 is Northern dispalcemt of A from C
so use sin again to find angle opp = 1.405 and hyp = 9.4
gives sin = .01495 so angle is 8.6 degrees
add on 270 for true bearing = 278.6